On the dimension of linear spaces of nilpotent matrices

Linear Algebra and its Applications 436 (2012) 2210–2230

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On the dimension of linear spaces of nilpotent matrices G.W. MacDonald a,∗,1 , J.A. MacDougall b , L.G. Sweet a a b

Department of Mathematics and Statistics, University of Prince Edward Island, Charlottetown, PE, Canada C1A 4P3 School of Mathematical and Physical Sciences, University of Newcastle, Callaghan, NSW 2308, Australia

ARTICLE INFO

ABSTRACT

Article history: Received 2 March 2011 Accepted 19 October 2011 Available online 25 November 2011

We obtain bounds on the dimension of a linear space S of nilpotent n × n matrices over an arbitrary field. We consider the case where bounds k and r are known for the nilindex and rank, respectively, and find the best possible dimensional bound on the subspace S in terms of the quantities n, k and r. We also consider the case where information is known concerning the Jordan forms of matrices in S and obtain new dimensional bounds in terms of this information. These bounds improve known bounds of Gerstenhaber. Along the way, we generalize a result of Mathes, Omladiˇc, and Radjavi concerning traces on subspaces of nilpotent matrices. This is a key component in the proof of our result and may also be of independent interest. © 2011 Elsevier Inc. All rights reserved.

Submitted by R.A. Brualdi AMS classification: Primary: 15A30 15B48 Secondary: 47D03 Keywords: Matrix Subspace Nilpotent Maximal dimension Rank Nilindex

1. Introduction In has been half a century since Murray Gerstenhaber established the seminal results concerning algebras and subspaces of nilpotent matrices over arbitrary fields. In first paper in the sequence [2], in 1958, Gerstenhaber showed that if S is a subspace of the vector space of n × n matrices over some field F, S consists of nilpotent matrices, and the field F is sufficiently large, then the maximal dimension of n(n−1) . He also gave a characterization of those cases where the maximal dimension is attained. S is 2 In the last paper in the sequence [3], in 1962, Gerstenhaber gave an improved bound on the dimension of S in terms of possible sizes of Jordan blocks in the Jordan forms of matrices in S. ∗ Corresponding author. E-mail addresses: [email protected] (G.W. MacDonald), [email protected] (J.A. MacDougall), [email protected] (L.G. Sweet). 1 The author acknowledges the support of NSERC Canada. 0024-3795/$ - see front matter © 2011 Elsevier Inc. All rights reserved. doi:10.1016/j.laa.2011.10.028

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There has been only moderate progress in this area since. In 1985, Serežkin [6] removed the cardinality condition in showing that Gerstenhaber’s 1958 results are valid over any field. In 1991, Mathes, Omladiˇc and Radjavi [5] gave new and quite elementary proofs of both Gerstenhaber’s 1958 results and Serežkin’s generalizations. In 1993, Brualdi and Chavey [1] used combinatorial methods to obtain a dimensional bound on S in the case where all matrices in S have nilindex bounded by k. They also used these methods to give a combinatorial proof of a result of Gerstenhaber [3] concerning the case where all matrices in S have rank bound r. In 2009, the second and third authors of this paper considered a case where conditions were placed on both the rank and the nilindex [7], and showed that if the maximal nilindex is two and the rank is bounded by r, then the dimension of S is bounded by r (n − r ). In this paper we make two substantial contributions to the general problem of obtaining dimension bounds for subspaces of nilpotent matrices. First, we obtain a sharp bound on the dimension of a subspace of n × n nilpotent matrices with rank bound r and nilindex bound k, in terms of the quantities k, r and n. Secondly, we give an advance over Gerstenhaber’s 1962 Theorem by giving an improved bound on the dimension of S in terms of data obtained from the possible Jordan forms of matrices in S. The only background knowledge required is of standard theorems and results from linear algebra. Before proceeding, we review some standard definitions and terminology which will be used throughout the paper. For a field F, we let char (F) denote the characteristic of the field and card(F) denote the cardinality ∗ of the field. We denote by Fn , the dual space of Fn , the usual n-dimensional vector space over F. We let Mn (F) denote the n × n matrices over F, Mmn (F) denote the m × n matrices over F and Un (F) denote the set of all strictly upper triangular matrices in Mn (F). As usual, for a matrix A ∈ Mn (F), we say A is nilpotent if some power of A equals 0, and define the nilindex of A to be the smallest natural number k so that Ak = 0. Also, let tr (A) denote the trace of A and let AT denote the transpose of A. We also adopt the convention that all matrices shall be denoted by capital letters and their entries  by  the corresponding lower-case letters. So if A is in Mn (F), its (i, j)-th entry is aij and we write A = aij . If V is a vector space over F, we let dim(V ) denote its dimension. For any rational number x, x, the floor function of x, is the largest integer which is less than or equal to x, and   x, the ceiling function of x, is the smallest integer which is greater than or equal to x. We define

0 0

= 1.

In Section 2, we give a new proof of a lemma of Mathes, Radjavi and Omladiˇc [5] concerning a simple trace condition that must be satisfied by any space of nilpotent matrices and generalize it to a slightly larger class of fields. We then state and prove a Dimension Slicing Lemma (which is a slight generalization of Lemma 1 of [7]) which will figure prominently in many of our proofs. Finally we apply our generalization of the Mathes, Radjavi and Omladiˇc Lemma and our Dimension Slicing Lemma to recast the elementary proof in [5] of Gerstenhaber’s 1958 Theorem over arbitrary fields in terms of these elements and an orthogonality argument. These techniques are core to obtaining our new results and so we include it to give the reader some familiarity with some of these techniques before seeing them in the more advanced setting. In Section 3, we obtain a block-matrix generalization of the Theorem of Mathes, Omladiˇc and Radjavi. It gives additional trace conditions which must be satisfied by subspaces of nilpotent matrices, and takes into account the maximum nilindex of matrices in the space. It will be a key component in the sharpening the dimensional bounds in our main theorems, and should also be of independent interest. In Section 4, we establish a number of technical lemmas required to prove our main theorems. In Section 5, we use our block-matrix generalization of the Theorem of Mathes, Omladiˇc and Radjavi from Section 3, our Dimension Slicing Lemma from Section 2, technical lemmas from Section 4, and induction on n (the size of the matrices) to prove our first major dimension-bounding theorem: we establish the bound on the dimension of a space of nilpotent matrices in terms of the nilindex bound k and the rank bound r. We show that, if S is a subspace of Mn (F), card(F) > n, and S consists of nilpotent matrices with nilindex bound k and rank bound r, then

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dim(S )  where q

=



 nr − r

k−1



r2 2

−

r 2

+

q2 2

q

(k − 1) + (−2r + k − 1) 2

.

A number of known results, including some of Brualdi and Chavey [1], are special cases of this result and are obtained as corollaries. In Section 6, we show that our bound is sharp by exhibiting examples of maximal dimension in all feasible cases. We then compare and contrast best previously known dimensional bounds on subspaces of nilpotent matrices with our result. In Section 7, we define a spatial Jordan partition of a subspace of nilpotent matrices. This is a list of numbers determined by considering possible Jordan forms of matrices in our subspace and is related to the definition of Jordan partition used by Gerstenhaber. Again using our block-matrix generalization of the Theorem of Mathes, Omladiˇc and Radjavi from Section 3, our Dimension Slicing Lemma from Section 2 and our technical lemmas from Section 4, we obtain improved dimensional bounds in terms of this quantity. Section 7 is independent of Sections 5 and 6 and so readers interested mainly in the results in that section may go directly to that section after Section 4. While Gerstenhaber used techniques and theorems from algebraic geometry, and Brualdi and Chavey used techniques and theorems from combinatorics, all of our proofs use only linear algebra.

2. The Mathes–Omladiˇc–Radjavi Theorem Following is the result that appears as Corollary 1 in [5]. Theorem 2.1 (Mathes, Omladiˇc, and Radjavi). Suppose S is a linear space of nilpotent matrices over a field of characteristic 0. If A, B ∈ S and k ∈ N then tr (Ak B) = 0. We now provide a very elementary proof of a slightly stronger result. Theorem 2.2. Suppose S is a subspace of Mn (F) and S consists of nilpotent matrices. If A, B k ∈ N and card(F)  n and k + 1 is not a multiple of char (F) then tr (Ak B) = 0

∈ S and

= 0 so the result is obvious fork  n. Assume  1  k  n − 1. For ∈ F, xA + B is nilpotent so tr (xA + B)k+1 = 0. But tr (xA + B)k+1 is a polynomial in x of degree at most k (since the coefficient of xk+1 is tr (Ak+1 ) which is zero since Ak+1 is nilpotent) and since card(F)  n > k the coefficient of each term must be 0. In particular the coefficient of xk is

Proof. Since A is nilpotent, An any x

tr (Ak B + Ak−1 BA + Ak−2 BA2

+ · · · + BAk ) = (k + 1)tr (Ak B) = 0.

Every field is an integral domain, so if k + 1 must have tr (Ak B) = 0 as required. 

= 0 in the field (i.e. k + 1 is not a multiple of char (F)) we

Corollary 2.3. Suppose S is a subspace of Mn (F) and S consists  ofcommuting nilpotent matrices. If k+l A, B ∈ S, k, l ∈ N, card(F)  k + l, and the binomial coefficient k is not a multiple of char (F), then tr (Ak Bl )

= 0. 

Proof. The proof is similar to the above. Consider tr



(xA + B)k+l . 

We can weaken the hypotheses of Theorem 2.1 even further by removing the condition about the char (F).

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Theorem 2.4. Suppose S is subspace of Mn (F) and S consists of nilpotent matrices. If A, B and card(F) > k then tr (Ak B) = 0.

∈ S, k ∈ N

Proof. Using a similarity transformation we may assume that A is in Jordan form. We may also assume 1  k  n − 1. For B in S and x ∈ F, the characteristic polynomial of xA + B is cxA+B (λ) =  λn = ni=0 cn−i (x)λi . The coefficient ck+1 (x) is identically zero and is the sum of the (k + 1) × (k + 1) principal minors of xA + B. This is a polynomial of degree at most k in x whose leading term is ± tr (Ak B). Since card(F) > k the coefficient of each term must equal 0, so tr (Ak B) = 0.  In the remainder of this paper we use a Dimension Slicing Lemma, which is a slight generalization of a lemma proved by two of the authors in a previous paper [7]. Given a subspace S of Mmn (F), and a set of indices I in {1, 2, . . . , m} × {1, 2, . . . , n}, we associate two subspaces of Mmn (F) to S. The first is WS , which is constructed by taking all the elements of S and “zeroing out” the entries whose index is not in I. So

aij if (i, j) ∈ I . WS = W ∈ Mmn (F) : there exists A ∈ S , with wij = 0 if (i, j)  ∈ I The second subspace is  US = S ∈ S : sij



= 0 whenever (i, j) ∈ I .

Lemma 2.5 (Dimension Slicing Lemma). For S a subspace of Mmn (F), I a subset of {1, 2, . . . , m} {1, 2, . . . , n}, and WS and US defined as above, we have that dim(S )

×

= dim(WS ) + dim(US ).

Proof. Consider the linear transformation P aij if (i, j) ∈ I, bij = 0 if (i, j)  ∈ I.

: Mmn (F) → Mmn (F) defined by P (A) = B where

Restrict P to S and apply the Rank-Nullity Theorem [4]. Clearly the range of P |S is WS and the kernel of P |S is US , so the result follows.  Using Theorem 2.4 and Lemma 2.5 we can recast the proof in [5] and give a short proof of Serežkin’s extension of Gerstenhaber’s Theorem [2] for arbitrary fields. Theorem 2.6 (Gerstenhaber/Serežkin). Suppose S is a subspace of Mn (F) and S consists of nilpotent n(n−1) . matrices, then dim(S )  2 Proof. Let I = {(i, j) : 1  j < i  n}. Then WS is the space spanned by all lower triangular parts of the elements of S, and US is the space consisting of all the matrices in S for which all the entries below the diagonal are 0. Since all the matrices in S are nilpotent, it follows that US actually consists of all the matrices in S which are strictly upper triangular, so US is a subspace of Un (F). Now WST , the transpose of WS , is also a subspace of Un (F) and is isomorphic to WS . Define a non-singular inner product on Un (F) by A, B = tr (ABT ). Theorem 2.4 implies that WST is contained in the orthogonal complement of US and so using Lemma 2.5: dim(S )

= dim(WS ) + dim(US ) = dim(WST ) + dim(US )  dim (Un (F)) =

n(n − 1) 2

.



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Note that we only used the conclusion of Theorem 2.4 when k field.

= 1 so the result is valid for any

3. A new trace condition for nilpotent spaces In this section we derive a new condition which generalizes Theorem 2.1 in a different direction. It gives additional trace conditions for spaces of nilpotent matrices and again the proof is elementary. First, we need a few definitions. Definition 3.1. The diagonals of an r × s matrix A are the sets  Dt = aij : 1  i  r , 1  j  s, j − i = t for 1 − r  t  s − 1. Thus D0 is the set of entries on the main diagonal of A, D1 is the set of entries on the first superdiagonal of A, D−1 is the set of entries on the first subdiagonal of A, and so on, until Ds− 1 = {a1s } and D1−r = {ar1 }. The trace of an r × s matrix is the sum of the entries in D0 . Also, we let Di denote the sum of all of the elements in Di . In the following theorem we make extensive use of the Jordan normal form of a nilpotent matrix (see [4]). In particular, if S is a space of nilpotent matrices we choose a matrix J in S with the largest possible nilindex k = k1 and then use a similarity transformation to assume that J is transformed to its Jordan normal form. Then J k = 0 but J k−1  = 0 and J = ⊕Jki where each Jki is the ki × ki matrix ⎡

J ki

=

0 0 0 ⎢ ⎢ ⎢1 0 0 ⎢ ⎢ ⎢0 1 0 ⎢ ⎣.

..

···

..

.

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ k i × ki

(which we refer to as a ki × ki Jordan block) and where k = k1  k2  k3 . . . (If J contains a zero block in the lower righthand corner we consider that block as a direct sum of 1 × 1 zero blocks.) If A is any matrix in S, we partition A into blocks of sizes corresponding to the sizes of the Jordan blocks Ji . So the block Aij is of size ki × kj . If B = J α AJ β and B is partitioned to correspond to A, then it is easy to show that, since J is block diagonal, Bij

β

= Jiα Aij Jj .

Theorem 3.1. Suppose S is a subspace of Mn (F), S consists of nilpotent matrices and card(F)  n. Let Q be any matrix in S of maximal nilindex k = k1 and let J = ⊕Ji be the Jordan normal   form of Q , where Ji is a ki × ki matrix. If A is any matrix in S and A is written as a block matrix A = Aij (with respect to the same decomposition as Q ), then tr (Aij Jjl )

= 0 for all l such that 0  k − ki  l  kj − 1.

Proof. Let x ∈ F. Then (J + xA)k = 0 and since card(F)  n all the coefficients of the matrix polynomial (J + xA)k must equal 0. The coefficient of x gives us J k−1 A + J k−2 AJ

+ J k−3 AJ 2 + · · · + AJ k−1 = 0.

Each entry of this sum provides a linear relationship on the entries of A. Now let B = (J + xA)k and partition B as described above. Then Bij

= Jik−1 Aij + Jik−2 Aij Jj + Jik−3 Aij Jj2 + · · · + Aij Jjk−1 = 0.

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If Ji  = 0 then multiplying a matrix (of the appropriate size) on the left by Ji has the effect of shifting each row of that matrix down one row, and adding a row of zeros at the top of the matrix. Similarly, if Jj  = 0 then multiplying a matrix on the right by Jj has the effect of shifting the columns of the matrix one column to the left and adding a column of zeros in the rightmost column of the matrix. Now consider the entry in the lower-left corner of Bij . Using the expansion for Bij above, we see that k−l

l−1

it is a sum of k terms. Each of these terms is the lower-left entry of a Ji Aij Jj for l = 1, 2, . . . , k. Thus it is obtained from an entry of Aij which, after being shifted down k − l times and shifted to the left l − 1 times, ends up in the lower-left corner. Looking at it from a different perspective, these are the entries of Aij that can be reached by starting in the lower left corner and doing a total of k − 1 moves either to the right or upwards. These are precisely the entries of Aij in Dk−ki (possibly with additional zeros if these moves can take us outside the index bound of the matrix Aij ). We can do a similar argument for each entry in the last row of Bij and see that the last row of Bij is 

Dk−ki ,



Dk−ki +1 , . . . ,



 Dkj −1 , 0, 0, . . . , 0

= 0,

where there are k − ki zeroes at the end of the row. If k = ki then there are no zeroes, and we obtain that the sum along each diagonal of Aij above and including the main diagonal is zero. If k − ki > kj − 1 (so k > (ki + kj ) − 1) then the last row is all kj = k or Jj = 0 and zeroes and we obtain no information. Also the argument is still valid if Ji = 0 and  ki = k but in these cases we only determine that the (1, kj ) entry of Bij , which is Dkj −1 is 0. Informally, this means that the sum of each of the (ki + kj ) − k highest diagonals of Aij are zero. Finally we note that this result can also be expressed as tr (Aij Jjl )

= 0 for all l such that 0  k − ki  l  kj − 1.

(In the case where l

= 0, we follow the usual convention that Jj0 is the kj × kj identity matrix.) 

4. Technical lemmas In this section we prove a number of technical lemmas required to prove our main theorems. A key step will be to use our Dimension Slicing Lemma, Lemma 2.5 to break S into “diagonal” and “off-diagonal” subspaces. The main difficulty will be in bounding the dimension of the “off-diagonal” subspaces. For this we will again use Lemma 2.5, as well as Theorem 3.1 and some clever calculating. That is the content of the following technical lemma. n n k n−k . For each A ∈ Mn (F), identify A For ⎤ n, decompose F as a direct sum F = F ⊕ F ⎡1  k  A11 A12 ⎦ , where A11 is k × k, A12 is k × (n − k), A21 is (n − k) × k and A22 is (n − k) × (n − k). with ⎣ A21 A22 Lemma 4.1. Suppose S is a subspace of Mn (F) so that S consists of nilpotent matrices of bounded nilindex k, and there exists N ∈ Mn−k,n−k (F) so that ⎡ ⎤ 0 J k ⎦∈S J=⎣ 0 N (where Jk is a k × k Jordan block). If card(F)  n, then the subspace ⎧⎡ ⎫ ⎡ ⎤ ⎤ ⎨ 0 A ⎬ A A 12 ⎦ : there exists A = ⎣ 11 12 ⎦ ∈ S X = ⎣ ⎩ A ⎭ A21 A22 0 21

has dim(X )

 (k − 1)(n − k).

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G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

Proof. With no loss of generality, we may assume that J is in Jordan form with first block Jk , since it can be transformed into such via a block diagonal similarity which will not change the dimension of X. Let A ∈ S and let x ∈ F. Then (J + xA)k = 0 and since card(F)  n all the coefficients of the matrix polynomial (J + xA)k must equal 0. The coefficient of x2 gives us D

= J k−2 A2 + J k−3 A2 J + J k−3 AJA + · · · + A2 J k−2 = 0.

This sum consists of all words of length k containing k − 2 J’s and two A’s. Each entry of this sum provides a quadratic relationship on the entries of A. As before, multiplying a matrix on the left by J has the effect of shifting the rows of the matrix down one row and adding a row of zeroes at the top of the matrix, and multiplying a matrix on the right by J has the effect of shifting columns of the matrix one column to the left and adding a column of zeroes in the rightmost column. In each term in the above formula for D there are always k − 2 shifts. Now consider the entry dk1 . This is simply the sum of the dot product of the kth row and the first column of each term of the above expression, if each term is written as a product of two matrices. Let ⎡ ⎤ A1 R ⎣ ⎦ A= C A2 be the block matrix of A with respect to the above decomposition. So A1 is k × k and A2 is (n − k)×(n − k). Now C is (n − k) × k, and we denote its columns by c1 , c2 , . . . , ck . These are vectors in F(n−k) . Also, R is (n − k) × k, and we denote its rows by r1T , r2T , . . . , rkT , so r1 , r2 , . . . , rk are also vectors in F(n−k) . Since J is block diagonal, each term of dk1 is a sum of terms of two types: products of pairs of entries from A1 and products of pairs of entries, with one from R and one from C. Furthermore, terms which are products of pairs of entries from A1 must be of the form aij alm where one of the two terms is in the strictly lower triangular part of A1 , and one is in the upper triangular part of A1 . To see this, consider such a pair, and with no loss of generality assume that aij comes from the first A1 and alm from the second A1 . The aij must be moved to the last row and so originates from a term in the above expansion k−i A1 . Similarly, alm must be moved to the first column and so originates from m−1 . The total number of Jk ’s in any term is k − 2 and so the aij alm terms comes a term that ends in A1 Jk k−i i−m−1 m−1 i−m−1 from the term Jk A1 Jk A1 Jk . We can consider the effect of Jk in the middle as either k−1 shifting the last row of Jk A1 (which is [ai1 , ai2 , . . . , aik ]) to the left (and inserting zeros at the end) m−1 (which is [a1m , a2m , . . . , akm ]T ) down and adding zeros at the or shifting the first column of A1 J1 k−1 i−m−1 m−1 A1 Jk ) we will top. Considering it as the latter, when we compute the (k, 1) entry of (Jk A1 )(Jk

for D that begins with Jk

be multiplying

 T [ai1 , ai2 , . . . , aik ] 0, . . . , 0, a1m , a2m , . . . , a(k−i+m+1)m ((i − m − 1) zeros were inserted at the top of the column). Thus, when we compute this product, a1m is matched with ai(i−m) , a2m is matched with ai(i−m+1) and in general alm is matched with aij where j = i − m + l − 1. (If any index goes out of bounds that terms is considered to be 0.) We can now see that alm comes from the upper triangular part of A1 if and only if l  m if and only if j  i − 1 < i if and only if aij comes from the strictly lower triangular part of A1 . So with no loss of generality we may assume that these terms are of the form aij alm where i > j and l  m. We claim that, in dk1 , the sum over all products of pairs of entries from A1 is zero; that is, dk1 is actually a sum of products of entries, with one from R and one from C. To see this, consider all terms in dk1 which contain a given aij where i > j. Such a term arises when aij is shifted into the kth row or the first column. The first condition occurs when there is a word in the expression for D of the form (J k−i A)B where B is a word of length i − 1 containing exactly one A

G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

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and i − 2 J’s. The second condition occurs when there is a word in the expression for D of the form B(AJ j−1 ) where B is a word of length k − j containing exactly one A and k − j − 3 J’s. By taking all possible valuesof B for the two cases it can be observed that the sum of all of the terms containing aij is actually ai, ( D(i−j)−1 ). Since (i − j) − 1  0, Theorem 3.1 gives us that D(i−j)−1 = 0 . What terms from R and C appear in the expression for dk,1 ? No entry from r1T (the first row of R) or ck (the last column of C) can appear, since it would take k − 1 shifts to move these terms to the last row of R or first column of C and we only have k − 2 shifts available. Note that from the term J k−2 A2 we obtain r2T c1 since the second row of R is shifted to the kth row and then multiplied by the first column of C. From the term J k−3 A2 J we obtain r3T c2 since it is the third row of R which is shifted (k − 3) times to the kth row and it is the second column of C which is shifted to the first column. Similarly, for each term of the form J k−i A2 J i−2 , for i = 2, 3, . . . , k − 2 we get a contribution riT ci−1 . We can never obtain in this expression an entry from an ith row of R times an entry from a jth column of C where j − i > 1, since it would take k − i down shifts to move the entries of riT to the kth row and j − 1 left shifts to move the jth column of C to the first column but the total number of down shifts plus left shifts available is at most k − 2. However, terms of the form J k−i AJ l AJ j−2 where (j − i) + l = 0 can contribute terms which include an entry of an ith row of R and j column of C where j − i < 1. If we let ⎡ ⎤ ⎡ ⎤ r c ⎢ 2⎥ ⎢ 1 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ r3 ⎥ ⎢ c2 ⎥ ⎢ ⎥ ⎢ ⎥ r = ⎢ . ⎥ and c = ⎢ . ⎥ ⎢.⎥ ⎢ . ⎥ ⎢.⎥ ⎢ . ⎥ ⎣ ⎦ ⎣ ⎦ rk ck−1 then these are vectors in F(k−1)(n−k) and the equation dk,1 = 0 can be expressed as r T Xc = 0, where X is an (k − 1)(n − k) × (k − 1)(n − k) lower triangular matrix with ones on the diagonal and thus is invertible. The exact formula for the matrix X is ⎡ ⎤ I 0 ··· 0 ⎢ n−k ⎥ ⎢ ⎥ ⎢ N In−k 0 ⎥ ⎢ ⎥ ⎥, X =⎢ . .. ⎢ . ⎥ . ⎢ . ⎥ ⎣ ⎦ k−2 k−3 N N In−k where the block entry in each block on the ith subdiagonal is N i , so X depends only on the matrix J and is independent of the matrix A. We now apply Lemma 2.5 to X , using the index set I = {k + 1, k + 2, . . . , n} × {1, 2, . . . , k}. Then ⎧⎡ ⎫ ⎤ ⎨ 0 0 ⎬ ⎦ ⎣ WX = : there exists A ∈ S with A21 = C ⎭ , ⎩ C 0

UX

=

⎧⎡ ⎤ ⎡ ⎤ ⎨ 0 R 0 R ⎦:⎣ ⎦ ⎣ ⎩ 0 0 0 0

⎫ ⎬

∈S , ⎭

and dim(X )

= dim(WX ) + dim(UX ).

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Consider the map T1

∗

: UX → F(k−1)(n−1) defined by

⎡ ⎤ ⎡ 0 R ⎦ ⎣ 0 0

=

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0

and the map T2

r1T

···

⎤

⎥ ⎥ r2T ⎥ ⎥ .. ⎥ ⎥ .⎥ ⎥ ⎥ rkT ⎥ ⎦ 0 0 0

: WX → F(k−1)(n−k) defined by

⎡ ⎤ 0 0 ⎦ ⎣ C 0 ⎡

=

T1

−→ [r2 , r3 , · · · , rk ] = r

0

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ c1

···

⎤

⎥ ⎥ 0⎥ ⎥ .. ⎥ ⎥ .⎥ ⎥ ⎥ 0⎥ ⎦ ck−1 ck 0

⎡ T2

−→

⎢ ⎢ ⎢ ⎢ ⎣

⎤ c1

⎥

.. ⎥ ⎥ . ⎥ = c. ⎦

ck−1

Both these maps are injective, by Theorem 3.1, as each entry in r1 (resp. ck ) is the negative of a sum of entries in r2 , . . . , rk (resp. c1 , . . . , ck−1 ). Hence dim(T1 (UX )) = dim(UX ) and dim(T2 (WX )) = dim(W ⎤⎞ ⎤ ⎤ ⎛⎡ ⎡ ⎡ ⎡X ). ⎤

0 R 0 R 0 R+Q 0 Q ⎦⎠, q = ⎦ ∈ X then ⎣ ⎦ ∈ UX and ⎣ ⎦ ∈ X . Letting r = T1 ⎝⎣ If ⎣ 0 0 C 0 0 0 C 0 ⎤⎞ ⎤⎞ ⎛⎡ ⎛⎡ 0 Q 0 0 ⎦⎠ and c = T2 ⎝⎣ ⎦⎠, we obtain that (r + q) Xc = 0. But we have rXc = 0, so for T1 ⎝⎣ 0 0 C 0 ∗

all q ∈ T1 (UX ) and c ∈ T2 (WX ), we have qXc = 0. So T1 (UX ) in F(k−1)(n−k) is perpendicular to XT2 (WX ) in F(k−1)(n−k) . Putting it all together (using that X is invertible) we obtain that dim(X ) = dim(UX ) + dim(WX )

= dim(T1 (UX )) + dim(T2 (WX )) = dim(T1 (UX )) + dim(XT2 (WX ))  dim(F(k−1)(n−k) ) = (k − 1)(n − k) and the lemma is proven.  In Section 5, we will show that if S is a subspace of Mn (F) (where card(F) > n) and S consists of nilpotent matrices whose nilindex is less than or equal to k and whose rank less than or equal to r, then dim(S )  where q

=

 nr −



r

k−1



r2 2

−

r 2

+

q2 2

q

(k − 1) + (−2r + k − 1) 2

.

We shall be using induction on n to prove this, but it will make it easier to verify the induction step if we know, apriori, that our dimension bound for S, br ,n (k)

= nr −

r2 2

−

r 2

+

q2 2

q

(k − 1) + (−2r + k − 1) 2

G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

2219

is an increasing function   of the nilindex bound k (holding the rank bound r and the matrix size n fixed). Recall that q

=

r k−1

, so this assertion is not immediately obvious. It is true nonetheless, and this is

a direct consequence of the following lemma. Lemma 4.2. For r d(k)

=

∈ N, the function !2

r k−1

(k − 1) +

is an increasing function of k for k Proof. For a fixed k

Let m

=

r

k(k−1)

k−1

(−2r + k − 1)

= 2, 3, . . . , r + 1.

∈ {2, 3, . . . , r }, we must show

d(k + 1) − d(k) is non-negative. 

!

r

=

r k

 . Then r

!2

(k) +

r k

!

(−2r + k) −

r k−1

!2

(k − 1) −

r k−1

!

(−2r + k − 1)

∈ M = [mk(k − 1), (m + 1)k(k − 1)) and

{Ii = [mk(k − 1) + ik, mk(k − 1) + (i + 1)k) : i = 0, 1, . . . , k − 2} and 

Jj



= [mk(k − 1) + j(k − 1), mk(k − 1) + (j + 1)(k − 1)) : j = 0, 1, . . . , k − 1

are two partitions of M. It is easy to see that Ii ∩ Jj = ∅ unless either j = i or j = i + 1, so we must have that either: (1) r ∈ Ii ∩ Ji for some i = 0, 1, 2, . . . , k − 1; or (2) r ∈ Ii ∩ Ji+1 for some i = 0, 1, 2, . . . , k − 1. In case (1) we have that mk(k − 1) + ik  r < mk(k − 1) + (i + 1)(k − 1) and r

!

k

= m(k − 1) + i,

r

!

k−1

= mk + i.

In case (2) we have that mk(k − 1) + (i + 1)(k − 1) r k

!

= m(k − 1) + i,

r k−1

!

 r < mk(k − 1) + (i + 1)k and

= mk + i + 1.

Substituting these values into d(k + 1) − d(k) and using the bound for r we obtain that in case (1) d(k + 1) − d(k)

 m2 (k2 − k) + 2m(ki) + (i2 + i),

while in case (2) d(k + 1) − d(k)

 m2 (k2 − k) + 2m ((k − i) + (mi − 1)) + (i2 + i).

In case (1), it is immediate that d(k + 1) − d(k)  0. In case (2), by considering the formula in the subcases where m = 0 or i = 0 separately we also obtain that d(k + 1) − d(k)  0.  5. Dimensional bounds from nilindex and rank bounds We now state and prove our result obtaining dimensional bounds from nilindex and rank bounds.

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G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

Theorem 5.1. Suppose S is a subspace of Mn (F) and S consists of nilpotent matrices whose nilindex  is r , less than or equal to k and whose rank is less than or equal to r. If card(F) > n then, setting q = k− 1 we have that

dim(S )

 nr −

r2 2

−

r 2

+

q2 2

q

(k − 1) + (−2r + k − 1). 2

Proof. As mentioned, we shall prove this induction on n. The base case is n = 1. Clearly, in this case, S must be the zero subspace, so r = 0 and k = 1. The bound expression clearly simplifies to zero so the theorem is true when n = 1. Next suppose the theorem is true for all dimensions 1, 2, 3, . . . , n − 1 and that S is as in the theorem. By Lemma 4.2 there is no loss of generality in assuming that there is a matrix Q in S which achieves the nilindex bound k, and with no loss of generality assume Q is in its Jordan form (as a similarity applied to S can achieve this without changing dimension, maximal nilindex or rank bound). So Q = ⊕Jki where each Jki is a Jordan block and the first Jordan block Jk1 = Jk is the largest. Write Q in 2 × 2 block matrix form ⎤ ⎡ Jk 0 ⎦. ⎣ Q = 0 N Choose two matrices A and B in S. With respect to this same decomposition, we express A in S as a 2 × 2 block matrix ⎡ ⎤ A11 A12 ⎦, A=⎣ A21 A22 where A11 ∈ Mkk (F), A22 ∈ M(n−k)(n−k) (F), A21 ∈ M(n−k)k (F), and A12 ∈ Mk(n−k) (F). Consider the index set I = {1, 2, . . . , k} × {k + 1, k + 2, . . . , n} ∪ {k + 1, k + 2, . . . , n} × {1, 2, . . . , k}, and for this index set define WS and US as in Lemma 2.5. Then WS is a subspace of Mn (F) with zeros in the (1, 1) and (2, 2) blocks, and US is the set of all the matrices in S of the form ⎡ ⎤ A1 0 ⎣ ⎦. 0 A2 By Lemma 2.5, dim(S )

= dim(WS ) + dim(US )

and by Lemma 4.1 dim(WS )

 (k − 1)(n − k).

To find dim(US ), we again apply Lemma 2.5 to US with the index set I = {1, 2, . . . , k} × {1, 2, . . . , k}. Then we obtain WUS

=

and U US

=

⎧⎡ ⎤ ⎨ A 0 1 ⎦ ⎣ ⎩ 0 0

⎤

⎡

: there exists A2 so that ⎣

⎧⎡ ⎤ ⎡ ⎤ ⎨ 0 0 0 0 ⎣ ⎦:⎣ ⎦ ⎩ 0 A 0 A2 2

⎫ ⎬

∈ S⎭ .

A1 0 0 A2

⎦

∈S

⎫ ⎬ ⎭

G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

2221

× k nilpotent matrices with no other rank or nilindex restrictions so the

The elements of WUS are k

(k−1)k

by Theorem 2.6. Also, UUS is essentially a set of (n − k) × best bound we can get is dim(WUS )  2 (n − k) nilpotent matrices, and so is amenable to our induction hypothesis. Summarizing, we now have that dim(S ) = dim(WS ) + dim(US )

(1)

 (k − 1)(n − k) + dim(WUS ) + dim(UUS ) k(k − 1)  (k − 1)(n − k) + + dim(UUS ).

(2) (3)

2

The dimension of UUS is clearly equal to the dimension of ⎧ ⎫ ⎡ ⎤ ⎨ ⎬   ⎣0 0 ⎦ ∈ U US ⎭ . S = A :  ⎩ 0 A We apply our ⎡ 0 n. If A = ⎣ 0

induction hypothesis to this⎡subspace S  . Clearly the matrix size is n ⎤ ⎤ 0 J 0 ⎦ ∈ UU , then for Q = ⎣ k ⎦ as above, A + xQ is in S for any x S  A 0 N

= n−k < ∈ F and so

k − 1 + rank(xN + A )  r. Thus rank(xN + A )  r − k + 1 for all x ∈ F, which implies that rank(A )  r − (k − 1) = r + 1 − k for all A ∈ S  . So the new rank bound we use applying the induction hypothesis to UUS will be r  = r + 1 − k. Clearly, k , the maximal nilindex of matrices A ∈ S  can be no more than k (since k is the maximal index for S), but we must also have k  n − k or there would not be room for a second k × k Jordan block. Also since the rank of a Jordan block of size k is k − 1, we must have (k − 1) + (k − 1)  r or we would exceed the allowable rank, so k  r + 2 − k. Thus k , the new maximal nilindex of S  will satisfy 1

 k  min{k, n − k, r + 2 − k}.

Since Lemma 4.2 gives that our dimension formula increases as a function of k, with no loss of   = min{k, n − k, r + 2 − k}. Let q = kr−1 .

generality we may assume k

Consider the following three comprehensive possibilities.

Case 1: k =  k  min  {n− k, r +2 − k} 1−k Here, q = r + = k−r 1 − 1 = q − 1. k−1 Our induction hypotheses gives that dim(S  )

 n r  −

r

2

−

2

r 2

+

q

2

2

 (n − k)(r + 1 − k) − +

(q − 1)

2

2

(k − 1) +

(k − 1) +

q 2

(r + 1 − k)2 2

(q − 1) 2

(−2r  + k − 1) −

(r + 1 − k) 2

(−2r − 2 + 2k + k − 1).

Then dim(S )

= dim(WS ) + dim(WUS ) + dim(S  )  (k − 1)(n − k) +

(k − 1)k 2

+ (n − k)(r + 1 − k) −

(r + 1 − k)2 2

2222

G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

−

(r + 1 − k)

= nr − Case 2: k Here q

=

2 r2

−

2

r

+ +

2

(q − 1)2 q

2

2

2

(k − 1) +

(q − 1) 2

(−2r − 2 + 2k + k − 1)

q

(k − 1) + (−2r + k − 1). 2

=  r + 2 −k  n − k < k r +1−k = 1. r +2−k−1

Our induction hypothesis gives that dim(S  )

 n r  −

r

2

2

−

r 2

q

+

2

2

 (n − k)(r + 1 − k) − +

(k − 1) +

q 2

(r + 1 − k)2 2

(−2r  + k − 1) −

(−2r − 2 + 2k + r + 2 − k − 1)

(r + 1 − k) 2

+

(r + 1 − k) 2

2

 (n − k)(r + 1 − k) −

(r + 1 − k)2 2

+

(−r + k − 1) 2

.

Then, dim(S ) = dim(WS ) + dim(WUS ) + dim(S  )

 (k − 1)(n − k) +  nr − But since r

r2 2

−

3r 2

(k − 1)k 2

−

= nr −

r2 2 r2 2

− −

3r 2 r 2

(r + 1 − k)2 2

+

(−r + k − 1) 2

+ k − 1.

+ 2 − k < k, we have that

dim(S )  nr

+ (n − k)(r + 1 − k) −

r k−1

< 2 and so q = 1 in this case. Thus

+k−1 +

q2 2

q

(k − 1) + (−2r + k − 1). 2

Case 3: k = n − k < min{k, r + 2 − k} In this case, we have that n − k < r + 2 − k. This implies that r + 2 > n and this is only possible if Q consists of a single n × n Jordan block. Thus we have that the rank bound for S is r = n − 1 and the maximal nilindex is k = n. In this degenerate case, r  = 0, n = 0 and k = 1. So dim(S  ) = 0 and we obtain that dim(S ) = dim(WS ) + dim(WUS ) + dim(S  )

 (k − 1)(n − k) + =

n(n − 1) 2

(k − 1)k 2

+0

.

It is easily verified that for r = n − 1 and k = n, our formula collapses to the Gerstenhaber bound from Theorem 2.6 as well, so in this case we also obtain that dim(S )

 nr −

r2 2

−

r 2

+

q2 2

q

(k − 1) + (−2r + k − 1).

This completes the proof of Theorem 5.1. 

2

G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

2223

Suppose S is a subspace of Mn (F), which consists of nilpotents of bounded rank r, but we are given no additional information on the maximal nilindex k. By considering Jordan forms, it is clear that all we can say is that the largest the nilindex can be is k = r + 1. As our bound in Theorem 5.1 increases with k, we obtain the following corollary as a special case Theorem 5.1, which recaptures a result of Gerstenhaber [3]. Corollary 5.2 [1,3]. Suppose S is a subspace of Mn (F) and S consists of nilpotent matrices of rank less than or equal to r. If card(F) > n then dim(S )

 nr −

r2 2

r

− . 2

Proof. By the comments preceding the Corollary, this follows by setting k the bound in Theorem 5.1. 

= r + 1 (and so q = 1) in

Another special case of note is when our subspace consists of square-zero matrices. In this case our Theorem recovers the result of [7]. Corollary 5.3 [7]. Suppose S is a subspace of Mn (F) and S consists of square-zero matrices of rank less than or equal to r. If card(F) > n then dim(S )

 nr − r 2 .

Proof. Apply Theorem 5.1, with k

= 2. 

One last special case that deserves consideration is when S is a subspace of Mn (F), which consists of nilpotents of maximal nilindex k, but we are given no additional information concerning a bound on the rank of matrices in S. By considering the Jordan form of matrices " # in S, and noting that each Jordan

block is at most k × k, we determine that there must be at least

n k

Jordan blocks in the Jordan form.

(For a real number x, x is the ceiling function of x, i.e. the smallest integer which is greater than or equal to x.) Each"block of a matrix in S # has one-dimensional kernel, so the dimension of the kernel " # must be at least

n k

. Hence, the ranks of matrices in S are bounded by r

= n−

n k

. This gives the

following cumbersome corollary. Corollary 5.4. Suppose S is a subspace of Mn (F) and S consists of nilpotent matrices of maximal nilindex k. If card(F) > n then $ % &' % &'2 % &' $ $ 1 1 n n n − − dim(S )  n n − n− n− k 2 k 2 k " # ⎥⎞ " # ⎥⎞ ⎛⎢ ⎛⎢ ⎢n − n ⎥ 2 ⎢n − n ⎥ $ $ ' % &' 1 ⎢ 1 ⎢ n k ⎥ k ⎥ ⎦⎠ (k − 1) + ⎝⎣ ⎦⎠ −2 n − + ⎝⎣ +k−1 . 2 k−1 2 k−1 k Proof. By the comments preceding the Corollary, this follows by applying Theorem 5.1, with " #⎥ ⎢ ⎢n − n ⎥ % & ⎢ n k ⎥ ⎦.  r =n− and q = ⎣ k k−1 In the case where k divides n, this formula simplifies significantly. Corollary 5.5. Suppose S is a subspace of Mn (F) and S consists of nilpotent matrices of maximal nilindex k. If k divides n and card(F) > n then

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G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

dim(S )

 n2

$

k−1

'

2k

.

Proof. In this case, the bound for the rank is r values gives the result. 

= n − nk , so q = nk . Applying Theorem 5.1 with these

The above two Corollaries consider a case contained in the paper of Brualdi and Chavey [1]. Corollary 3.6 of that paper says the following. Corollary 5.6 [1]. Let k be an integer with 2  k  n. Let W be a linear space of nilpotents in Mn (F), each with index at most k. If F is sufficiently large, then ⎞ ⎛ k  1 2 2 dim(W )  ⎝n − ci ⎠ , 2 i=1 where γ

= (c1 , c2 , . . . , ck ) is a partition of n into k parts with parts differing by at most 1.

The definition of partition of n used in Brualdi and Chavey [1] is a non-increasing sequence of natural numbers which sum to n. It is not too difficult  to see that, in the case where k divides n, the best bound is achieved by the constant partition γ = nk , nk , . . . , nk , in which case Corollary 5.6 gives the same bound as our Corollary 5.5. In the general case, it is possible that Corollary 5.6 gives the same bound as our Corollary 5.4.

6. Construction of nilpotent spaces of maximal rank In this section we show that our bound in Theorem 5.1 is sharp by constructing subspaces of nilpotent spaces of maximal dimension in all feasible cases. A feasible case is a triple (k, r , n) for which there exists a subspace Sk,r ,n in Mn (F) consisting of nilpotent matrices of maximal rank r, and maximal nilindex k. As mentioned previously, consideration of possible Jordan forms of matrices in Sk,r ,n leads immediately to the condition that 0  k − 1  r  n − 1. " #Also note that, for k is the maximal nilindex, each matrix A ∈ Sk,r ,n in Mn (F) must have at least n k

blocks in its Jordan form. With each block there corresponds a zero column and so A has at most " # " # n . n − nk non-zero columns. Thus, we obtain that r  n − nk , or equivalently k  n− r These are the only conditions on the triple (k, r , n) required for feasibility.

Theorem 6.1. For each triple (k, r , n) of non-negative integers satisfying n k r+1n n−r there exists a subspace Sk,r ,n in Mn (F), consisting of nilpotent matrices of maximal rank r, and maximal nilindex k and having dimension dim(Sk,r ,n ) = nr   r where q = k− . 1

−

r2 2

−

r 2

+

q2 2

Proof. Given n, k, r and q as above, let s Then decompose Fn as q terms

(

)*

+

2

= r − q(k − 1) + 1. So r = q(k − 1) + s − 1 where 0  s < k.

F = F ⊕ Fk ⊕ · · · Fk ⊕ Fs ⊕ Fn−r −q−1 . n

k

q

(k − 1) + (−2r + k − 1),

G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

2225 q+2

Let Sk,r ,n be the set of all matrices S in Mn (F) whose (q + 2) × (q + 2) block matrices [Sij ]i,j=1 with respect to the above decomposition, satisfy the following conditions: (1) For 1  i, j  q, Sij is strictly lower-triangular k × k matrix; (2) For 1  i  q, Si,q+1 is a k × s matrix which has its k highest diagonals (Ds−k , . . . , Ds−1 ) equal to zero; (3) For 1  i  q + 2, Si,q+2 = 0; (4) For 1  j  q, Sq+1,j has its s highest diagonals (Dk−s−1 . . . , Dk−1 ) equal to zero; (5) Sq+1,q+1 is a strictly lower triangular s × s matrix; (6) For 1  j  q, Sq+2,j is an (n − r − q − 1) × k matrix with last column consisting of zeroes; (7) Sq+2,q+1 is an (n − r − q − 1) × s matrix with last column consisting of zeroes. (It is possible that some of the blocks in the above decomposition are zero dimensional, in which case the corresponding blocks above are not present.) In general it is clear that the rank is bounded by r (since there will be n − r columns of zeroes), and this rank is achieved. It is slightly less obvious that the maximal nilindex is k. When we compute X 2 , all the (i, j) block entries move down or left creating more zeroes, except in the (q, q + 1) block where no new zero columns are created. But for each power after that, in every block a new column or diagonal of zeroes is created and since s < k, we obtain that X k = 0. Now we calculate the dimension of Sk,r ,n by computing the dimension of each block. Clearly each (i, j)th block, for 1

blocks, so these blocks contribute s(s−1) 2

 i, j  q, contributes

k(k−1) q2 2

k(k−1) 2

dimensions, and there are q2 such

total dimensions. The (q + 1, q + 1) block contributes

dimensions. The blocks in the (q + 2, j) entries contribute (n − r − q − 1)(r ) dimensions, since there are n − r − q − 1 entries in each column and there are r nonzero columns. By pairing off the (q + 1, i) entry with the (i, q + 1) entry, for i = 1, 2, . . . , q, we see that each paired block has exactly (k − 1)s arbitrary entries and there are q such blocks so they contribute q(k − 1)s dimensions. Thus dim(Sk,r ,n ) Substituting s

= q2

k(k − 1) 2

+

s(s − 1) 2

+ (n − r − q − 1)(r ) + q(k − 1)s.

= r − q(k − 1) + 1 and simplifying, we obtain the formula in Theorem 5.1. 

Example 6.1. If we apply Theorem 6.1, in the case where n = 12, r = 8 and k = 4, then S4,8,12 consists of all matrices X of the following form, where entries indicated by an ∗ are arbitrary elements of F: ⎡ ⎤ 0 0 0 0 0 0 0 0 0 0 0 0 ⎢ ⎥ ⎢ ∗ 0 0 0 ∗ 0 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∗ ∗ 0 0 ∗ ∗ 0 0 ∗ 0 0 0 ⎥ ⎢ ⎥ ⎢ ∗ ∗ ∗ 0 ∗ ∗ ∗ 0 ∗ ∗ 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 0 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ∗ 0 0 0 ∗ 0 0 0 0 0 0 0 ⎥ ⎥. X =⎢ ⎢ ⎥ ⎢ ∗ ∗ 0 0 ∗ ∗ 0 0 ∗ 0 0 0 ⎥ ⎢ ⎥ ⎢ ∗ ∗ ∗ 0 ∗ ∗ ∗ 0 ∗ ∗ 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∗ 0 0 0 ∗ 0 0 0 0 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ∗ ∗ 0 0 ∗ ∗ 0 0 ∗ 0 0 0 ⎥ ⎢ ⎥ ⎣ ∗ ∗ ∗ 0 ∗ ∗ ∗ 0 ∗ ∗ 0 0 ⎦

∗

∗

∗

0

∗

∗

∗

0

∗

∗

Theorem 5.1 and Theorem 6.1 give that dim(S4,8,12 ) number of ∗ in the above matrix.

0

0

= 53 which can be verified by counting the

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G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

If we only use the information about the rank bound and apply Corollary 5.2, we obtain a weaker bound that dim(S4,8,12 )  68, while if we only use information about the nilindex bound, and apply Corollary 5.5 (since k divides n), we obtain a improved but still less than optimal bound that dim(S4,8,12 )  54. Note that in [3], Gerstenhaber also provides a general bound for the dimension of a subspace S of nilpotent matrices, in terms of all the possible sizes of Jordan blocks which occur in the Jordan form of all the matrices in S. This would, most likely, be difficult to determine in any particular case. One advantage of our formula is that it depends only on a rank bound and a nilindex bound; information that may be more accessible for a subspace of nilpotents S than the possible sizes of blocks in all possible Jordan decompositions of matrices in S. In cases where information about the possible structure of Jordan forms of matrices in the subspace is available, we can again use our methods to give a dimensional bound, and this bound improves that of Gerstenhaber in many cases.

7. Improving Gerstenhaber’s General Theorem In order to state Gerstenhaber’s General Theorem, which relates the dimensional bound on a subspace of nilpotent matrices to information about the sizes of Jordan blocks in Jordan forms of matrices in S, we need some preliminary terminology. We use the notation and terminology of [1]. For n a positive integer, we say α = (a1 , a2 , . . . , an ) is a partition of n if a1  a2  · · · an  0, and n = a1 + a2 + · · · + a n . The conjugate of the partition α is the partition of n defined by α ∗ = (a∗1 , a∗2 , . . . , a∗n ) where a∗j is the number of ai in α which are greater than or equal to j. We define a partial order  on the set of partitions of n as follows: if α = (a1 , a2 , . . . , an ) and β = (b1 , b2 , . . . , bn ) are two partitions of n, then α  β if a1 kl

+ · · · + aj  b1 + · · · bj for all j = 1, 2, . . . , n.

To each nilpotent matrix A ∈ Mn (F) we associate a partition of n as follows: Let k1  k2  · · ·   1 be the sizes of the Jordan blocks in the Jordan form of A. Then n = k1 + k2 + · · · + kl and so jp(A)

= (k1 , k2 , . . . , kl , 0, . . . , 0)

is a partition of n (we have adjoined n − l zeros) which is called the Jordan partition of A. If S is a subspace of Mn (F) consisting of nilpotent matrices, theJordan partition of S is the least upper bound (in the partial order mentioned above) of the set of all Jordan partitions of matrices in S. Gerstenhaber’s General Theorem [3] is the following. Theorem 7.1 (Gerstenhaber). Suppose S is a subspace of Mn (F) consisting of nilpotent matrices, and that γ = (c1 , . . . , cn ) is the conjugate of the Jordan partition of S. If card(F) is sufficiently large, then ⎞ ⎛ n  1 2 2 ci ⎠ . dim(S )  ⎝n − 2 i=1 As this bound involves only possible sizes of Jordan blocks of matrices in S, it can still be quite coarse. Consider the following two examples. Example 7.1. For n an even positive integer, and k

S1

=

⎧⎡ ⎤ ⎨ X X 11 12 ⎣ ⎦ : X11 , X12 , X21 , X22 ⎩ X X 21 22

= 2n , let ⎫ ⎬

∈ Uk (F)

⎭

G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

2227

and let S2

=

⎧⎡ ⎤ ⎨ X X 11 12 ⎣ ⎦ : X11 , X12 , X21 , X22 ⎩ X X 21 22

⎫ ⎬

∈ Uk (F), X11 = X22 . ⎭

In both cases it is obvious that the maximal nilindex is k and so it immediately follows that jp(S1 ) = jp(S2 ) = (k, k, 0, . . . , 0) and the conjugate of this partition is γ = (2, . . . , 2, 0 . . . , 0) where there are k twos and k zeros. Thus, for i = 1, 2, Gerstenhaber’s General Theorem gives the bound of ⎛ ⎞ k  1 2 2 dim(Si )  ⎝n − 2 ⎠ 2 i=1 ' $ 1 n n(n − 2) for i = 1, 2. = n2 − (4) = 2 2 2 From direct computation is it easy to see that this bound is sharp in the case of S1 , but the actual dimension of S2 is three quarters of this bound. We offer a finer bound which can distinguish between subspaces which have the same Jordan partition by introducing a spatial component. Definition 7.1. For S a subspace of Mn (F) consisting of nilpotents, we define a spatial Jordan partition of S to be

ζ = (k1 , k2 , . . . , kl ), where k1

 k2  · · · kl  1 are inductively defined as follows:

(1) k1 is the maximal nilindex of matrices in S. Fix an A1 in S which has this maximal index, and let Fn = M1 ⊕ N1 , where A1 has block matrix ⎡ ⎤ J k1 0 ⎣ ⎦ 0 N with respect to this decomposition. (Or equivalently, apply a similarity transformation to our subspace S so that A has the form above.) Let S1 be the subspace of all matrices in S whose block matrix with respect to this decomposition is of the form ⎤ ⎡ 0 0 ⎦. ⎣ 0 X (2) Inductively define ki+1 , Mi+1 , Ni+1 and Si+1 as follows: choose Ai+1 ∈ Si of maximal nilindex ki+1 in Si . Let Ni = Mi+1 ⊕ Ni+1 , so that, with respect to the decomposition Fn = (M1 ⊕ · · · ⊕ Mi ) ⊕ Mi+1 ⊕ Ni+1 , Ai has block matrix ⎡

0

0

0

⎤

⎢ ⎥ ⎢ ⎥ ⎢0 Jki+1 0 ⎥ . ⎣ ⎦ 0 0 N (Or equivalently, apply a similarity transformation to our subspace S using an invertible matrix of the form I ⊕ I · · ·⊕ I ⊕ T, where I’s are used on blocks corresponding to Jk1 , . . . , Jki .)

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Then let Si+1 be the subspace of all matrices in S whose block matrix with respect to this decomposition is of the form ⎤ ⎡ 0 0 0 ⎥ ⎢ ⎥ ⎢ ⎢0 0 0 ⎥ . ⎦ ⎣ 0 0 X Note that, if at any point, Si = {0} then kj = 1 for all j = i + 1, . . . , l, as these correspond to one-dimensional Jordan blocks (i.e. zero matrices). We are now ready to state our general theorem providing dimensional bounds on subspaces of nilpotent matrices in terms of spatial Jordan partitions. Theorem 7.2. Suppose S is a subspace of Mn (F) consisting of nilpotent matrices, and that ζ is a spatial Jordan partition of S. If card(F) > n then ⎛ ⎞ l i   ki dim(S )  (ki − 1) ⎝ + n − kj ⎠ . 2 i=1 j=1

= (k1 , . . . , kl )

Proof. The proof requires only our Dimensional Slicing Lemma (Lemma 2.5), our lemma for bounding off diagonal terms (Lemma 4.1) and the dimensional bound for a general subspace of nilpotents (Theorem 2.6). Let Ai and Si , for i = 1, 2, . . . , l, be as in Definition 7.1. With no loss of generality (by applying a similarity), we may assume that A1 is in its Jordan form, with Jordan blocks arranged in order of decreasing size. So, with respect to the decomposition Fn =

Fk1 ⊕ Fn−k1

⎤

⎡ A1

=⎣

J k1 0 0 N

⎦.

We proceed similarly to the proof of Theorem 5.1. Consider the index set I = {1, 2, . . . , k1 } × {k1 + 1, k1 + 2, . . . , n} ∪ {k1 + 1, k1 + 2, . . . , n} × {1, 2, . . . , k1 }, and for this index set define WS and US as in Lemma 2.5. Then WS is a subspace of Mn (F) with zeros in the (1, 1) and (2, 2) blocks, and US is the set of all the matrices in S of the form ⎤ ⎡ 0 A 1 ⎦ ⎣

.

0 A2

By Lemma 2.5, dim(S )

= dim(WS ) + dim(US )

and by Lemma 4.1 dim(WS )

 (k1 − 1)(n − k1 ).

To find dim(US ), we again apply Lemma 2.5 to US with the index set I = {1, 2, . . . , k1 } × {1, 2, . . . , k1 }. Then we obtain WUS

=

⎧⎡ ⎤ ⎨ A 0 1 ⎣ ⎦ ⎩ 0 0

⎤ 0 A 1 ⎦ there exists A2 so that ⎣ 0 A2 ⎡

:

⎫ ⎬

∈ S⎭

G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

2229

and U US

=

⎧⎡ ⎤ ⎡ ⎤ ⎨ 0 0 0 0 ⎣ ⎦:⎣ ⎦ ⎩ 0 A 0 A2 2

The elements of WUS are k1 dim(WUS )



⎫ ⎬

∈S . ⎭

× k1 nilpotent matrices, so by Theorem 2.6,

(k1 − 1)k1 2

.

Now, UUS is a set of (n − k1 ) × (n − k1 ) nilpotent matrices, which is our set S1 . So dim(S ) = dim(WS ) + dim(US )

 (k1 − 1)(n − k1 ) + dim(WUS ) + dim(UUS ) k1 (k1 − 1)  (k1 − 1)(n − k1 ) + + dim(S1 ). 2

We now repeat this argument to bound dim(S1 ) in terms of the new maximal nilindex k2 , the new dimension n − k1 and the dimension of S2 to obtain dim(S )  (k1

− 1)(n − k1 ) +

 (k1 − 1)(n − k1 ) + +

k2 (k2

− 1)

2

k1 (k1

− 1)

2

k1 (k1

− 1)

2

+ dim(S1 ) + (k2 − 1)(n − k1 − k2 )

+ dim(S2 ).

Repeating l times and simplifying we obtain the required bound.  In Example 7.1 it is not to difficult to see that a spatial Jordan partition of S1 is ζ1 = (k, k) while a spatial Jordan partition of S2 is ζ2 = (k, 1, . . . , 1) (there are k ones). For the first subspace, our Theorem 7.2 gives the same bound as Gerstenhaber’s General Theorem dim(S1 )



n(n − 2) 2

but for the second subspace, our Theorem 7.2 gives the improved bound dim(S2 )



3n(n − 2) 8

which is sharp. Informally, it could be said that Gerstenhaber’s General Theorem is unable to identify and account for multiplicity in the Jordan forms of matrices in the subspace, while our bound takes multiplicity into account and so it should give a bound which is at least as good as Gerstenhaber’s in all cases. At present we are unable to determine if this is the case but are willing to make the following conjecture. Conjecture 7.3. Suppose S is a subspace of Mn (F) consisting of nilpotent matrices and card(F ) > n. Then the dimensional bound in Theorem 7.2 is less than or equal to the dimensional bound in Theorem 7.1. Serežkin eventually improved Gerstenhaber’s Theorem to remove any condition on the underlying field while retaining the same bound on the dimension of the subspace of nilpotent matrices. Is it possible the hypothesis that card(F) > n could be dropped in Theorem 7.2 and still retain the dimensional

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G.W. MacDonald et al. / Linear Algebra and its Applications 436 (2012) 2210–2230

bound in the conclusion? This is an open problem. We know of no counterexample in small fields, but do not have enough evidence or intuition to conjecture in either direction. References [1] [2] [3] [4] [5] [6] [7]

R.A. Brualdi, K.L. Chavey, Linear spaces of toeplitz and nilpotent matrices, J. Combin. Theory Ser. A 63 (1993) 65–78. M. Gerstenhaber, On nilalgebras and varieties of nilpotent matrices I, Amer. J. Math. 80 (1958) 614–622. M. Gerstenhaber, On nilalgebras and varieties of nilpotent matrices IV, Ann. of Math 75 (1962) 382–418. K. Hoffman, R. Kunze, Linear Algebra, second ed., Prentice Hall, 1971. B. Mathes, M. Omladiˇc, H. Radjavi, Linear spaces of nilpotent matrices, Linear Algebra Appl. 149 (1991) 215–225. V.N. Serežkin, Linear transformations preserving nilpotency, Vest¯i Akad. Navuk BSSR Ser. F¯iz.-Mat. Navuk 5 (1985) 46–50. L.G. Sweet, J.A. MacDougall, The maximum dimension of a subspace of matrices of index 2, Linear Algebra Appl. 8 (2009) 1116–1124.